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3 years ago

A quadratic function has a vertex at (2,5) and passes through the point (4,9).

1 answer:

4 0

Since another way to write an equation for a parabola is a(x-h)^2-k, with h and k as the vertex points, we can plug that in to get a(x-2)^2-5. To find a, we plug 4 in and have 9 as what it equals, getting a(4-2)^2-5=9=4a-5=9 and 4a=14, where a=14/4. Therefore, the equation would be (14/4)(x-2)^2-5

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Consider the line integral Z C (sin x dx + cos y dy), where C consists of the top part of the circle x 2 + y 2 = 1 from (1, 0) t

pashok25 [27]

Direct computation:

Parameterize the top part of the circle $x^2+y^2=1$ by

$\vec r(t)=(x(t),y(t))=(\cos t,\sin t)$

with $0\le t\le\pi$ , and the line segment by

$\vec s(t)=(1-t)(-1,0)+t(2,-\pi)=(3t-1,-\pi t)$

with $0\le t\le1$ . Then

$\displaystyle\int_C(\sin x\,\mathrm dx+\cos y\,\mathrm dy)$

$=\displaystyle\int_0^\pi(-\sin t\sin(\cos t)+\cos t\cos(\sin t)\,\mathrm dt+\int_0^1(3\sin(3t-1)-\pi\cos(-\pi t))\,\mathrm dt$

$=0+(\cos1-\cos2)=\boxed{\cos1-\cos2}$

Using the fundamental theorem of calculus:

The integral can be written as

$\displaystyle\int_C(\sin x\,\mathrm dx+\cos y\,\mathrm dy)=\int_C\underbrace{(\sin x,\cos y)}_{\vec F}\cdot\underbrace{(\mathrm dx,\mathrm dy)}_{\vec r}$

If there happens to be a scalar function $f$ such that $\vec F=\nabla f$ , then $\vec F$ is conservative and the integral is path-independent, so we only need to worry about the value of $f$ at the path's endpoints.

This requires

$\dfrac{\partial f}{\partial x}=\sin x\implies f(x,y)=-\cos x+g(y)$

$\dfrac{\partial f}{\partial y}=\cos y=\dfrac{\mathrm dg}{\mathrm dy}\implies g(y)=\sin y+C$

So we have

$f(x,y)=-\cos x+\sin y+C$

which means $\vec F$ is indeed conservative. By the fundamental theorem, we have

$\displaystyle\int_C(\sin x\,\mathrm dx+\cos y\,\mathrm dy)=f(2,-\pi)-f(1,0)=-\cos2-(-\cos1)=\boxed{\cos1-\cos2}$

3 0

3 years ago

Negative 5 times a number plus 8 is greater than the number minus 20

DanielleElmas [232]

Answer:

-5x+8 > x-20

Plz give brainiest

5 0

2 years ago

Read 2 more answers

How do you create and solve equations in one variable given a real–world situation?

WINSTONCH [101]

Step-by-step explanation:

1) first eliminate constant(number without variable) you eliminate constant by doing the inverse operation(if constant is being added, then subtract)(if constant is being subtracted, then add it)

ex: if x+9=32

subtract 9 from both sides to get rid of it

x=23

2) if there is another constant but this constant is being multiplied/divided then do the opposite operation to remove it

you could multiply by 3 on both sides:

x+9=69

x=60

5 0

2 years ago

A bag of 30 chocolate bars is shared equally among 12 children. which equation could be used to find the number of chocolate bar

IgorC [24]

Let the amount of chocolate bars each child receives be x

X= 30/12
X=5/2

8 0

3 years ago

Travel Agency offers vacation packages. Each vacation package includes a city, a month, and an airline. The agency has 3 cities,

ivanzaharov [21]

3 is the correct answer I believe

4 0

2 years ago